.TH std::ranges::rotate_copy,std::ranges::rotate_copy_result 3 "2024.06.10" "http://cppreference.com" "C++ Standard Libary"
.SH NAME
std::ranges::rotate_copy,std::ranges::rotate_copy_result \- std::ranges::rotate_copy,std::ranges::rotate_copy_result

.SH Synopsis
   Defined in header <algorithm>
   Call signature
   template< std::forward_iterator I, std::sentinel_for<I> S,

             std::weakly_incrementable O >
   requires std::indirectly_copyable<I, O>                           \fB(1)\fP \fI(since C++20)\fP
   constexpr rotate_copy_result<I, O>

       rotate_copy( I first, I middle, S last, O result );
   template< ranges::forward_range R, std::weakly_incrementable O >

   requires std::indirectly_copyable<ranges::iterator_t<R>, O>       \fB(2)\fP \fI(since C++20)\fP
   constexpr rotate_copy_result<ranges::borrowed_iterator_t<R>, O>

       rotate_copy( R&& r, ranges::iterator_t<R> middle, O result );
.SH Helper types
   template< class I, class O >                                      \fB(3)\fP \fI(since C++20)\fP
   using rotate_copy_result = in_out_result<I, O>;

   1) Copies the elements from the source range [first, last), to the destination range
   beginning at result in such a way, that the element *middle becomes the first
   element of the destination range and *(middle - 1) becomes the last element. The
   result is that the destination range contains a left rotated copy of the source
   range.
   The behavior is undefined if either [first, middle) or [middle, last) is not a valid
   range, or the source and destination ranges overlap.
   2) Same as \fB(1)\fP, but uses r as the source range, as if using ranges::begin(r) as
   first and ranges::end(r) as last.

   The function-like entities described on this page are niebloids, that is:

     * Explicit template argument lists cannot be specified when calling any of them.
     * None of them are visible to argument-dependent lookup.
     * When any of them are found by normal unqualified lookup as the name to the left
       of the function-call operator, argument-dependent lookup is inhibited.

   In practice, they may be implemented as function objects, or with special compiler
   extensions.

.SH Parameters

   first, last - the source range of elements to copy from
   r           - the source range of elements to copy from
   middle      - the iterator to the element that should appear at the beginning of the
                 destination range
   result      - beginning of the destination range

.SH Return value

   {last, result + N}, where N = ranges::distance(first, last).

.SH Complexity

   Linear: exactly N assignments.

.SH Notes

   If the value type is TriviallyCopyable and the iterator types satisfy
   contiguous_iterator, implementations of ranges::rotate_copy usually avoid multiple
   assignments by using a "bulk copy" function such as std::memmove.

.SH Possible implementation

   See also the implementations in libstdc++ and MSVC STL.

struct rotate_copy_fn
{
    template<std::forward_iterator I, std::sentinel_for<I> S, std::weakly_incrementable O>
    requires std::indirectly_copyable<I, O>
    constexpr ranges::rotate_copy_result<I, O>
        operator()(I first, I middle, S last, O result) const
    {
        auto c1 {ranges::copy(middle, std::move(last), std::move(result))};
        auto c2 {ranges::copy(std::move(first), std::move(middle), std::move(c1.out))};
        return {std::move(c1.in), std::move(c2.out)};
    }

    template<ranges::forward_range R, std::weakly_incrementable O>
    requires std::indirectly_copyable<ranges::iterator_t<R>, O>
    constexpr ranges::rotate_copy_result<ranges::borrowed_iterator_t<R>, O>
        operator()(R&& r, ranges::iterator_t<R> middle, O result) const
    {
        return (*this)(ranges::begin(r), std::move(middle),
                       ranges::end(r), std::move(result));
    }
};

inline constexpr rotate_copy_fn rotate_copy {};

.SH Example


// Run this code

 #include <algorithm>
 #include <iostream>
 #include <iterator>
 #include <vector>

 int main()
 {
     std::vector<int> src {1, 2, 3, 4, 5};
     std::vector<int> dest(src.size());
     auto pivot = std::ranges::find(src, 3);

     std::ranges::rotate_copy(src, pivot, dest.begin());
     for (int i : dest)
         std::cout << i << ' ';
     std::cout << '\\n';

     // copy the rotation result directly to the std::cout
     pivot = std::ranges::find(dest, 1);
     std::ranges::rotate_copy(dest, pivot, std::ostream_iterator<int>(std::cout, " "));
     std::cout << '\\n';
 }

.SH Output:

 3 4 5 1 2
 1 2 3 4 5

.SH See also

   ranges::rotate  rotates the order of elements in a range
   (C++20)         (niebloid)
   ranges::copy
   ranges::copy_if copies a range of elements to a new location
   (C++20)         (niebloid)
   (C++20)
   rotate_copy     copies and rotate a range of elements
                   \fI(function template)\fP
